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I need help with the fourier transform of Bessel function of first kind on 2 dimensions.

$$G(w_1,w_2) = F[J_0(a\sqrt{x^2+y^2})]$$

where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.

I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.

Cowboy Trader
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  • The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin. – Andy Walls Nov 19 '18 at 19:58

1 Answers1

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From Bracewell, the 2D cartesian Fourier Transform

$$F(u,v) = \mathscr{F}\left\{f(x,y)\right\} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) e^{-2\pi i (xu+yv)} dx dy$$

for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order

$$F(u,v) = F(q) = \mathscr{H}_0\left\{f(r)\right\} = 2\pi\int_{0}^{\infty}f(r)J_0(2\pi qr)r\space dr$$

with a strictly reciprocal inverse transform

$$f(x,y) = f(r) = \mathscr{H}_0\left\{F(q)\right\} = 2\pi\int_{0}^{\infty}F(q)J_0(2\pi qr)q\space dq$$

Note: $r = \sqrt{x^2+y^2}$ and $q=\sqrt{u^2+v^2}$

$ $

Looking at the inverse transform of $F(q) = \dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)$, where $\delta()$ represents a 1 dimensional delta function:

$$\begin{align*}f(r) = \mathscr{H}_0\left\{\dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)\right\} &= 2\pi\int_{0}^{\infty}\dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)J_0(2\pi qr)q\space dq\\ \\ &= \dfrac{2\pi}{a}\dfrac{a}{2\pi}J_0\left(2\pi \dfrac{a}{2\pi}r\right)\\ \\ &=J_0(ar)\\ \end{align*}$$

So that gives you your desired transform pair for real $a$:

$$\mathscr{F}\left\{J_0(ar)\right\} = \mathscr{H}_0\left\{J_0(ar)\right\} = \dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)$$

The transform is an impulse ring at a radius of $\dfrac{a}{2\pi}$ from the origin.

Andy Walls
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  • You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2\pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it. – Andy Walls Nov 19 '18 at 21:40
  • The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: https://en.m.wikipedia.org/wiki/Hankel_transform – Andy Walls Nov 19 '18 at 21:43
  • No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway. – Andy Walls Nov 19 '18 at 21:45
  • $|a|$ is always real and the result will hold for $a \ne 0$. $a=0$ is a special case that needs a separate transform. – Andy Walls Nov 19 '18 at 21:48